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Calculus 3 : Calculus 3

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Example Questions

Example Question #31 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(6\cdot (\frac{1}{4})^{(x^{2} + y^{2})})}{19}-\frac{ (23e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{3})dA\text{, where D is}\\&\text{the region defined between two circles centered on the origin with radii }\\&0.27\text{ and }1.18\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-1.000,0.031)\text{ and }\overrightarrow{u_2}=(0.992,-0.125)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-25.68

102.7

6.42

-4.28

Correct answer:

-25.68

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(6\cdot (\frac{1}{4})^{(x^{2} + y^{2})})}{19}-\frac{ (23e^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})})}{3})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(6\cdot (\frac{1}{4})^{(r^{2})}\cdot r)}{19}-\frac{ (23\cdot re^{(\frac{(2\cdot r^{2})}{3})})}{3})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.031}{-1.000})=0.99\pi;\theta_2=arctan(\frac{-0.125}{0.992})=1.96\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.99\pi}^{1.96\pi}\int_{0.27}^{1.18}(\frac{(6\cdot (\frac{1}{4})^{(r^{2})}\cdot r)}{19}-\frac{ (23\cdot re^{(\frac{(2\cdot r^{2})}{3})})}{3})drd\theta=(-\frac{ (23e^{(\frac{(2\cdot r^{2})}{3})})}{4}-\frac{ 3}{(19\cdot 4^{(r^{2})}ln(4))})d\theta|_{0.27}^{1.18}\\&\int_{0.99\pi}^{1.96\pi}(-8.425)d\theta=(-8.425\cdot \theta)|_{0.99\pi}^{1.96\pi}=-25.68\end{align*}$

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Example Question #32 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(34e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})} -\frac{ 3}{11})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.57\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.790,0.613)\text{ and }\overrightarrow{u_2}=(-0.094,-0.996)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

102.78

-4.28

-102.78

17.13

Correct answer:

17.13

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(34e^{(-\frac{ (3x^{2})}{2}-\frac{ (3y^{2})}{2})} -\frac{ 3}{11})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(34\cdot re^{(-\frac{(3\cdot r^{2})}{2})} -\frac{ (3\cdot r)}{11})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.613}{0.790})=0.21\pi;\theta_2=arctan(\frac{-0.996}{-0.094})=1.47\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.21\pi}^{1.47\pi}\int_{0}^{0.57}(34\cdot re^{(-\frac{(3\cdot r^{2})}{2})} -\frac{ (3\cdot r)}{11})drd\theta=(-\frac{ (34e^{(-\frac{(3\cdot r^{2})}{2})})}{3}-\frac{ (3\cdot r^{2})}{22})d\theta|_{0}^{0.57}\\&\int_{0.21\pi}^{1.47\pi}(4.328)d\theta=(4.328\cdot \theta)|_{0.21\pi}^{1.47\pi}=17.13\end{align*}$

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Example Question #41 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (19e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})})}{4}- 3sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.07\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.844,0.536)\text{ and }\overrightarrow{u_2}=(0.707,-0.707)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

8.9

-1.48

2.23

-8.9

Correct answer:

-8.9

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (19e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})})}{4}- 3sin(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2}))dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (19\cdot re^{(-\frac{(2\cdot r^{2})}{3})})}{4}- 3\cdot rsin(\frac{(3\cdot r^{2})}{2}))drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.536}{-0.844})=0.82\pi;\theta_2=arctan(\frac{-0.707}{0.707})=1.75\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\\&\int_{0.82\pi}^{1.75\pi}\int_{0}^{1.07}(-\frac{ (19\cdot re^{(-\frac{(2\cdot r^{2})}{3})})}{4}- 3\cdot rsin(\frac{(3\cdot r^{2})}{2}))drd\theta=(cos(\frac{(3\cdot r^{2})}{2}) +\frac{ (57e^{(-\frac{(2\cdot r^{2})}{3})})}{16})d\theta|_{0}^{1.07}\\&\int_{0.82\pi}^{1.75\pi}(-3.048)d\theta=(-3.048\cdot \theta)|_{0.82\pi}^{1.75\pi}=-8.9\end{align*}$

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Example Question #42 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(29e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})})}{2}-\frac{ cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}{33})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.38\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.187,0.982)\text{ and }\overrightarrow{u_2}=(-0.562,-0.827)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

10.66

-3.55

21.31

-85.25

Correct answer:

21.31

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(29e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})})}{2}-\frac{ cos(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}{33})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(29\cdot re^{(-\frac{(2\cdot r^{2})}{3})})}{2}-\frac{ (rcos(\frac{(2\cdot r^{2})}{3}))}{33})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.982}{0.187})=0.44\pi;\theta_2=arctan(\frac{-0.827}{-0.562})=1.31\pi\\&\text{Now, utilizing integral rules:}\\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.44\pi}^{1.31\pi}\int_{0}^{1.38}(\frac{(29\cdot re^{(-\frac{(2\cdot r^{2})}{3})})}{2}-\frac{ (rcos(\frac{(2\cdot r^{2})}{3}))}{33})drd\theta=(-\frac{ (87e^{(-\frac{(2\cdot r^{2})}{3})})}{8}-\frac{ sin(\frac{(2\cdot r^{2})}{3})}{44})d\theta|_{0}^{1.38}\\&\int_{0.44\pi}^{1.31\pi}(7.798)d\theta=(7.798\cdot \theta)|_{0.44\pi}^{1.31\pi}=21.31\end{align*}$

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Example Question #43 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(9e^{(2x^{2} + 2y^{2})} -\frac{ (5e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})}{6})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.61\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.969,0.249)\text{ and }\overrightarrow{u_2}=(0.827,-0.562)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

1114.53

-5572.65

-557.27

222.91

Correct answer:

1114.53

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(9e^{(2x^{2} + 2y^{2})} -\frac{ (5e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})}{6})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(9\cdot re^{(2\cdot r^{2})} -\frac{ (5\cdot re^{(-\frac{r^{2}}{2})})}{6})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.249}{-0.969})=0.92\pi;\theta_2=arctan(\frac{-0.562}{0.827})=1.81\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.92\pi}^{1.81\pi}\int_{0}^{1.61}(9\cdot re^{(2\cdot r^{2})} -\frac{ (5\cdot re^{(-\frac{r^{2}}{2})})}{6})drd\theta=(\frac{(9e^{(2\cdot r^{2})})}{4}+\frac{ (5e^{(-\frac{r^{2}}{2})})}{6})d\theta|_{0}^{1.61}\\&\int_{0.92\pi}^{1.81\pi}(398.6)d\theta=(398.6\cdot \theta)|_{0.92\pi}^{1.81\pi}=1114.53\end{align*}$

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Example Question #44 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(5e^{(- x^{2} - y^{2})})}{4}-\frac{ 4^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}}{20})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.82\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(-0.998,0.063)\text{ and }\overrightarrow{u_2}=(0.536,-0.844)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.03

-0.02

0.11

-0.23

Correct answer:

0.11

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(5e^{(- x^{2} - y^{2})})}{4}-\frac{ 4^{(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3})}}{20})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(5\cdot re^{(-r^{2})})}{4}-\frac{ (4^{(\frac{(2\cdot r^{2})}{3})}\cdot r)}{20})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.063}{-0.998})=0.98\pi;\theta_2=arctan(\frac{-0.844}{0.536})=1.68\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.98\pi}^{1.68\pi}\int_{0}^{1.82}(\frac{(5\cdot re^{(-r^{2})})}{4}-\frac{ (4^{(\frac{(2\cdot r^{2})}{3})}\cdot r)}{20})drd\theta=(-\frac{ (5e^{(-r^{2})})}{8}-\frac{ (3\cdot 4^{(\frac{(2\cdot r^{2})}{3})})}{(160ln(2))})d\theta|_{0}^{1.82}\\&\int_{0.98\pi}^{1.68\pi}(0.05161)d\theta=(0.05161\cdot \theta)|_{0.98\pi}^{1.68\pi}=0.11\end{align*}$

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Example Question #45 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3e^{(2x^{2} + 2y^{2})})}{8}-\frac{ (13\cdot (\frac{7}{2})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{3})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.05\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.976,0.218)\text{ and }\overrightarrow{u_2}=(0.790,-0.613)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-78.28

-39.14

234.85

7.83

Correct answer:

-39.14

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(3e^{(2x^{2} + 2y^{2})})}{8}-\frac{ (13\cdot (\frac{7}{2})^{(\frac{(3x^{2})}{2}+\frac{ (3y^{2})}{2})})}{3})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot re^{(2\cdot r^{2})})}{8}-\frac{ (13\cdot (\frac{7}{2})^{(\frac{(3\cdot r^{2})}{2})}\cdot r)}{3})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.218}{0.976})=0.07\pi;\theta_2=arctan(\frac{-0.613}{0.790})=1.79\pi\\&\text{Now, utilizing integral rules:}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.07\pi}^{1.79\pi}\int_{0}^{1.05}(\frac{(3\cdot re^{(2\cdot r^{2})})}{8}-\frac{ (13\cdot (\frac{7}{2})^{(\frac{(3\cdot r^{2})}{2})}\cdot r)}{3})drd\theta=(\frac{(3e^{(2\cdot r^{2})})}{32}-\frac{ (13\cdot (\frac{7}{2})^{(\frac{(3\cdot r^{2})}{2})})}{(9ln(\frac{7}{2}))})d\theta|_{0}^{1.05}\\&\int_{0.07\pi}^{1.79\pi}(-7.244)d\theta=(-7.244\cdot \theta)|_{0.07\pi}^{1.79\pi}=-39.14\end{align*}$

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Example Question #46 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(563e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})}{57})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.71\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.000,1.000)\text{ and }\overrightarrow{u_2}=(-0.998,-0.063)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-0.9

-14.38

3.59

1.8

Correct answer:

3.59

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(563e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})}{57})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(563\cdot re^{(-\frac{r^{2}}{2})})}{57})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{1.000}{0.000})=0.5\pi;\theta_2=arctan(\frac{-0.063}{-0.998})=1.02\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.5\pi}^{1.02\pi}\int_{0}^{0.71}(\frac{(563\cdot re^{(-\frac{r^{2}}{2})})}{57})drd\theta=(-\frac{(563e^{(-\frac{r^{2}}{2})})}{57})d\theta|_{0}^{0.71}\\&\int_{0.5\pi}^{1.02\pi}(2.201)d\theta=(2.201\cdot \theta)|_{0.5\pi}^{1.02\pi}=3.59\end{align*}$

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Example Question #47 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(\frac{5}{(\frac{2}{7})^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}})}{48}-\frac{ (29e^{(x^{2} + y^{2})})}{5})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&1.8\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.809,0.588)\text{ and }\overrightarrow{u_2}=(-0.368,-0.930)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

-261.71

-1308.57

87.24

1046.85

Correct answer:

-261.71

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(\frac{(\frac{5}{(\frac{2}{7})^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})}})}{48}-\frac{ (29e^{(x^{2} + y^{2})})}{5})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(5\cdot r)}{(48\cdot (\frac{2}{7})^{(\frac{r^{2}}{2})})}-\frac{ (29\cdot re^{(r^{2})})}{5})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.588}{0.809})=0.2\pi;\theta_2=arctan(\frac{-0.930}{-0.368})=1.38\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\\&\int_{0.2\pi}^{1.38\pi}\int_{0}^{1.8}(\frac{(5\cdot r)}{(48\cdot (\frac{2}{7})^{(\frac{r^{2}}{2})})}-\frac{ (29\cdot re^{(r^{2})})}{5})drd\theta=(-\frac{ (29e^{(r^{2})})}{10}-\frac{ 5}{(48\cdot (\frac{2}{7})^{(\frac{r^{2}}{2})}ln(\frac{2}{7}))})d\theta|_{0}^{1.8}\\&\int_{0.2\pi}^{1.38\pi}(-70.6)d\theta=(-70.6\cdot \theta)|_{0.2\pi}^{1.38\pi}=-261.71\end{align*}$

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Example Question #48 : Double Integration In Polar Coordinates

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(-\frac{ (3e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})}{7}-\frac{ (13e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{5})dA\text{, where D is}\\&\text{the region of a circle centered on the origin with a radius of }\\&0.64\\&\text{and encompassed by vectors from the origin }\\&\overrightarrow{u_1}=(0.218,0.976)\text{ and }\overrightarrow{u_2}=(0.613,-0.790)\\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Possible Answers:

0.67

-8.09

5.39

-2.7

Correct answer:

-2.7

Explanation:

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\\&\text{An approach that would be beneficial is a conversion to polar form:}\\&r=cos(\theta);r=sin(\theta)\\&r^2=x^2+y^2\\&dA=rdrd\theta\\&\text{With this we can find: }\iint_{D}(-\frac{ (3e^{(-\frac{ x^{2}}{2}-\frac{ y^{2}}{2})})}{7}-\frac{ (13e^{(\frac{x^{2}}{2}+\frac{ y^{2}}{2})})}{5})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(-\frac{ (3\cdot re^{(-\frac{r^{2}}{2})})}{7}-\frac{ (13\cdot re^{(\frac{r^{2}}{2})})}{5})drd\theta\\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.976}{0.218})=0.43\pi;\theta_2=arctan(\frac{-0.790}{0.613})=1.71\pi\\&\text{Now, utilizing integral rules:}\\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\\&\int_{0.43\pi}^{1.71\pi}\int_{0}^{0.64}(-\frac{ (3\cdot re^{(-\frac{r^{2}}{2})})}{7}-\frac{ (13\cdot re^{(\frac{r^{2}}{2})})}{5})drd\theta=(-\frac{(e^{(-\frac{r^{2}}{2})}\cdot (91e^{(r^{2})} - 15))}{35})d\theta|_{0}^{0.64}\\&\int_{0.43\pi}^{1.71\pi}(-0.6703)d\theta=(-0.6703\cdot \theta)|_{0.43\pi}^{1.71\pi}=-2.7\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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Example Question #31 : Double Integration In Polar Coordinates

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